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Let $x,y$ be two different transformations of order $2$ (i.e., they are reflections). Then we can form the two (abelian) groups $$(X,+)=\{0,x\} \quad \quad \text{and}\quad\quad(Y,+)=\{0,y\}$$ both of which are isomorphic to $\mathbb{Z}_2.$

Now define $$(G,+)=\{0,x\}\oplus \{0,y\}=\{0,x,y,x+y\},$$ where $\oplus$ is the direct sum. As far as I understand, $G$ is isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2.$ But wouldn't $$\mathbb{Z}_2\oplus\mathbb{Z}_2=\{0,1,0+1\}=\{0,1\}=\mathbb{Z}_2\;?$$

I guess the problem is that I can tell the difference between $x$ and $y$ in the case of $\{0,x\}\oplus \{0,y\}$ while I cannot in the case of $\mathbb{Z}_2\oplus\mathbb{Z}_2.$

This is probably elementary, but I'd really appreciate if someone would explain where I'm mistaken. Thanks.

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    Well, sure if you define $\oplus$ however you like, you can get your result. – 2017-02-25
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    @ThomasAndrews I wan't trying to do that. $\oplus$ is the direct sum - I take it to intuitively mean every possible combination of the elements of the two groups, where all combinations happen under the group operation (here $+$), so it behaves like a Cartesian product... Is this not standard? – 2017-02-25
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    Your expression of the set for $X\oplus Y $ is either not the definition or a poorly written one. Is that what your notes/book/reference says? – 2017-02-25
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    It is a Cartesian product and $(0,1) \neq (1,0)$ – 2017-02-25
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    @MarkS. No, it is rather based on [an answer](http://math.stackexchange.com/a/2157919/259085) to a question of mine. What would be an appropriate definition then be (when $x$ and $y$ commute)? – 2017-02-25
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    @PhillipHamilton Yes, for ordered sets (denoted by regular parentheses) that is true, but how does that apply to a group (a set (and an operation), denoted by curly brackets)? – 2017-02-25
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    @MarkS. Is it written more clearly now, or is it still borderline gibberish (I'm sorry if that is the case)? – 2017-02-25
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    @Lovsovs, I didn't mean to criticize you, just to say that your confusion could likely be solved by looking up a formal definition in a good textbook (and if you had already done so, I criticize that textbook), at least assuming you already understand cartesian products. You have some intuition from internal direct sums that is clouding your understanding of the external direct sum of groups that the notation $\mathbb Z_2 \oplus \mathbb Z_2$ signifies. – 2017-02-25
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    @MarkS. No worries, I didn't take it that way :) And thank you, I'll do that! – 2017-02-25

3 Answers 3

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Conside $G = ${$e, a, b, ab | a^2=e, b^2=e$}

Then $G$ is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2$

Let $f: G \to \mathbb{Z}_2 \oplus \mathbb{Z}_2$ by

$f(e) = (0,0) $

$f(a) = (1,0) $

$f(b) = (0,1) $

$f(ab) = (1,1) $

$f(ab) = (1,1) = (1,0) + (0,1) = f(a) + f(b)$

So $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ cannot be equal to $\mathbb{Z}_2$.

I think the issue is that your definition of Direct Sum is missing something important.

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    Would it then be a more appropriate notation to use $\mathbb{Z}_2\times\mathbb{Z}_2$, instead of the direct sum? Also (using the notation in your answer), how would you write out $\mathbb{Z}_2\oplus\mathbb{Z}_2,$ as you did with $G$? – 2017-02-25
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    @Lovsovs For the case of finitely many summands, the direct product and the direct sum are the same. – 2017-02-25
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The assertion that $$\mathbb{Z}_2 \oplus \mathbb{Z}_2 = \{0,1,0+1\} = \{0,1\}$$ is incorrect.

The idea of the direct sum of two groups is that we formally distinguish the elements of the first summand from those of the second, and apply the group operation to each component separately.

Formally, the (external) direct sum of two groups $(G,+_G)$ and $(H,+_H)$ is defined as the collection of pairs $$ G \oplus H = \{(g,h) \mid g \in G, h \in H\}$$ with group operation defined by $$ (g_1, h_1) +_{G\oplus H} (g_2, h_2) = (g_1 +_G g_2, h_1 +_H h2).$$

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    Thank you, that was very clear. – 2017-02-25
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    Btw, is $(G,+)=\{0,x\}\oplus \{0,y\}=\{0,x,y,x+y\},$ then not correct, when the group operations for the two summed groups are the same? If the group operation is the same, shouldn't their identities also be the same? – 2017-02-25
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    Even if the summands are the same group, we still distinguish the first and second components. The notation "$\{0,x,y,x+y\}$" is shorthand for "$\{0+0, x+0,0+y,x+y\}$", and elements "$0+0$", "$0+y$" etc. are themselves shorthand for $(0,0)$, $(0,y)$ etc. – 2017-02-25
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    In a direct sum, the element $(0,0,\dotsc,0)$ is always the identity, which you should be able to convince yourself of quite easily. :) – 2017-02-25
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We have: \begin{align} \mathbb Z_2\oplus\mathbb Z_2 &= \mathbb Z_2^{1\rm st}\oplus \mathbb Z_2^{2\rm nd}\\ &= \{0^{1\rm st},1^{1\rm st}\}\oplus\{0^{2\rm nd},1^{2\rm nd}\}\\ &= \{0^{1\rm st}+0^{2\rm nd}, 0^{1\rm st}+1^{2\rm nd}, 1^{1\rm st}+0^{2\rm nd}, 1^{1\rm st}+1^{2\rm nd}\} \end{align} Note that at this stage, the "$+$" notation in those elements is a mere convention, that will be justified afterwards. That is, we actually define the elements to be the complete expressions $x^{1\rm st}+y^{2\rm nd}$. Also note that addition is by definition component-wise, that is, $$(a^{1\rm st}+b^{2\rm nd}) + (c^{1\rm st}+d^{2\rm nd}) = (a+c)^{1\rm st}+(b+d)^{2\rm nd}$$

Now we see that $0^{1\rm st}+0^{2\rm nd}$ is the neutral element, so we just denote it as $0$. Moreover we notice that $\{0^{1\rm st}+x^{2\rm nd}:x\in \mathbb Z_2^{2\rm nd}\}$ is isomorphic to $\mathbb Z_2^{2\rm nd}$, therefore we omit the $0^{1\rm st}$, and analogously for $0^{2\rm nd}$.

Therefore we arrive at $$\mathbb Z_2\oplus\mathbb Z_2 = \{0, 1^{1\rm st},1^{2\rm nd}, 1^{1\rm st}+1^{2\rm nd}\}$$ Note however that $1^{1\rm st}\ne 1^{2\rm nd}$, therefore you cannot simplify this to $\{0,1,1,1+1\}=\{0,1\}$

Also note that now the "$+$" notation for the elements is justified after the fact: $x^{1\rm st}$, when standing alone, is just a shorthand for the expression $x^{1\rm st} + 0^{2\rm nd}$, and $y^{2\rm nd}$ is just a shorthand for the expression $0^{1\rm st}+y^{2\rm nd}$. But if we interpret $x^{1\rm st}+y^{2\rm nd}$ not as a single expression, but as the sum of the standalone values $x^{1\rm st} + 0^{2\rm nd}$ and $0^{1\rm st}+y^{2\rm nd}$, then we get $$(x^{1\rm st} + 0^{2\rm nd}) + (0^{1\rm st}+y^{2\rm nd}) = (x+0)^{1\rm st}+(0+y)^{2\rm nd} = x^{1\rm st}+y^{2\rm nd}$$ so we recover the original expression, which justifies the notation.

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    How can one formally show that $0^{1\rm st}+0^{2\rm nd}$ is actually the neutral element of the group formed by the direct sum? Also, how do we know that, for instance, $0^{1\rm st}$ acts neutrally on $1^{2\rm nd}$? – 2017-02-25
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    @Lovsovs: Because, by definition of the direct sum all elements of the direct sum are of the form $x^{1\rm st}+y^{2\rm nd}$ and addition is defined to be per component, therefore $(0^{1\rm st}+0^{2\rm nd}) + (x^{1\rm st}+y^{2\rm nd}) = (0+x)^{1\rm st}+(0+y)^{2\rm nd} = x^{1\rm st}+y^{2\rm nd}$, and same for the other order, which is exactly the definition of the neutral element. – 2017-02-25
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    @Lovsovs: I've extended the post a bit to clarify. – 2017-02-25
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    Thank you, showing the calculations made it much clearer. Note that I've edited some minor errors in your post: Please review the changes to make sure I didn't misinterpret the meaning. – 2017-02-25
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    Thank you for the corrections. As far as I can see, you didn't misinterpret anything. – 2017-02-25
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    Great! Btw, that component-wise definition of addition seems to be equivalent to vector-addition. What, exactly, is the deeper connection, if I may ask? – 2017-02-25
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    Because $\mathbb Z_2$ is a field, $\mathbb Z_2\oplus\mathbb Z_2$ indeed *is* a vector space over $\mathbb Z_2$ if you also define scalar multiplication component-wise. More generally, if you have a set of vector spaces over a common field (note that a field always is a vector space over itself), then their direct sum gives also a vector space over that field if you define scalar multiplication component-wise. BTW, thank you for the accept. – 2017-02-25
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    Beautiful, thank you! – 2017-02-25